Question

Researchers are interested in determining whether more men than women prefer Coca Cola to Pepsi. In a random sample of 300 men, 65% prefer Coca Cola, whereas in a random sample of 400 women, 48% prefer Coca Cola. What is the 99% confidence interval estimate for the difference between the percentages of men and women who prefer Coca Cola over Pepsi?

Answer 1

99% confidence interval for the difference between the percentages of men and women who prefer Coca Cola over Pepsi is between a lower limit of 13.3% and an upper limit of 20.7%.

Step-by-step explanation:

Confidence interval for a proportion is given as p +/- zsqrt[p(1-p) ÷ n]

difference between proportion (p) of men and women who prefer Coca Cola over Pepsi(p) = 0.65 - 0.48 = 0.17

total number of men and women (n) = 300+400 = 700

confidence level (C) = 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

critical value (z) at 1% significance level is 2.576.

Error of margin = zsqrt[p(1-p) ÷ n] = 2.576sqrt[0.17(1-0.17) ÷ 700] = 2.576 × 0.0142 = 0.037

Lower limit of difference in proportion = 0.17 - 0.037 = 0.133 = 13.3%

Upper limit of difference in proportion = 0.17 + 0.037 = 0.207 = 20.7%

99% confidence interval is (13.3%, 20.7%)